suppose the data represent the inches of rainfall in april for a certain city over the course of 20 years…

suppose the data represent the inches of rainfall in april for a certain city over the course of 20 years. given the quartiles q₁ = 2.400, q₂ = 3.945, and q₃ = 5.160, determine the lower and upper fences. are there any outliers, according to this criterion? the lower fence is - 1.740 (round to three decimal places as needed.) the upper fence is 9.300. (round to three decimal places as needed.) are there any outliers in the given data set? a. yes, there are values below the lower fence b. yes, there are values above the upper fence and below the lower fence. c. yes, there are values above the upper fence. d. no, all the values are between the lower and upper fences.

suppose the data represent the inches of rainfall in april for a certain city over the course of 20 years. given the quartiles q₁ = 2.400, q₂ = 3.945, and q₃ = 5.160, determine the lower and upper fences. are there any outliers, according to this criterion? the lower fence is - 1.740 (round to three decimal places as needed.) the upper fence is 9.300. (round to three decimal places as needed.) are there any outliers in the given data set? a. yes, there are values below the lower fence b. yes, there are values above the upper fence and below the lower fence. c. yes, there are values above the upper fence. d. no, all the values are between the lower and upper fences.

Answer

Explanation:

Step1: Recall outlier - fence formula

The lower fence is calculated as $Q_1 - 1.5\times(IQR)$ and the upper fence is calculated as $Q_3+1.5\times(IQR)$, where $IQR = Q_3 - Q_1$. Given $Q_1 = 2.400$, $Q_2=3.945$, $Q_3 = 5.160$. First, find $IQR$. $IQR=Q_3 - Q_1=5.160 - 2.400 = 2.760$

Step2: Calculate lower fence

Lower fence $=Q_1-1.5\times IQR=2.400-1.5\times2.760=2.400 - 4.140=-1.740$

Step3: Calculate upper fence

Upper fence $=Q_3 + 1.5\times IQR=5.160+1.5\times2.760=5.160 + 4.140 = 9.300$

Step4: Check for outliers

The data - set values range from $0.66$ to $6.13$. Since all values in the data - set ($0.66\leq x\leq6.13$) are between $-1.740$ and $9.300$, there are no outliers.

Answer:

D. No, all the values are between the lower and upper fences.